Question

14. The vertices of a 3ABC are A(-5 , 7), B(-4 , -5) and C(4 , 5). Find the slopes
of the altitudes of the triangle.

Answer 1

Let D , E and F are the points placed in sides BC, CA and AB of a triangle respectively in such a way that AD ⊥ BC , BE ⊥ CA and CF ⊥ AB.
Then, AD , BE and CF are known as altitudes of triangle ∆ABC .

so, To find equation of altitudes :
1. find slope of base in which it is perpendicular
Then, use formula, slope of altitude × slope of base = -1

Now, Let's do it ,
Slope of BC = (5 + 5)/(4 + 4) = 5/4
so, slope of AD × slope of BC = -1
∴ slope of AD = -4/5

Slope of CA = (5 - 7)/(4 + 5) = -2/9
slope of BE × slope of CA = -1
slope of BE = -1/(-2/9) = 9/2

slope of AB = (-5 - 7)/-4 + 5) = -12
slope of CF × slope of AB = -1
slope of CF = -1/(-12) = 1/12

Hence, slope of altitude AD = -4/5 , slope of altitude BE = 9/2 and slope of altitude CF = 1/12